# Show that it is measurable

Let $\mu$ a finite Borel measure in $\mathbb{R}^d$ and $B_x$ an open disc with center $x$ and radius $1$.

Show that the function $f(x)=\mu (B_x)$ is measurable.

Could you give me some hints how I could do that??

It's more than measurable, it's lower semicontinuous: Pick a sequence $x_k\to x$, then $1_{B_{x}}\leq \liminf_k 1_{B_{x_k}}$. Therefore by Fatou's lemma $$f(x)=\int_{B_x} d\mu \leq \int_{\mathbb{R}^n} \liminf_k 1_{B_{x_k}} d\mu \leq \liminf_k \int_{B_{x_k}} d\mu =\liminf_k f(x_k).$$
• Could you explain to me how we get the inequality $$1_{B_x} \leq \lim inf_k 1_{B_{x_k}}$$ ?? – Mary Star Feb 9 '15 at 19:59
• @MaryStar: If $y\in B_x$ then whenever we have $|x-x_k|<1-|x-y|$, we'll get $y\in B_{x_k}$. Arguing similarly if $y\notin \bar{B}_x$ we conclude that if $y\notin \partial B_x$ then $1_{B_{x_k}}(y)\to 1_{B_x}(y)$. But since, if $y\in \partial B_x$, $1_{B_x}(y)=0\leq 1_{B_{x_k}}(y)$, so the inequality follows. – Jose27 Feb 9 '15 at 22:33
• Could you explain it further to me?? And also when we have shown that $f(x) \leq \lim inf_k f(x_k)$ why does this mean that $f$ is semicontinuous?? – Mary Star Feb 10 '15 at 2:38
• Instead of showing that $f$ is semicontinuous, could we have also showed that $f$ is continuous?? Is the following correct?? When $x \rightarrow y$ we have that $f(x)-f(y) \rightarrow 0$ since the measure of the difference is the measure of the part of the balls that don't intersect. – Mary Star Feb 10 '15 at 2:46
Hint: Try $d=1$ first. Fubini's theorem might help.